https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Hansenhe&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T00:42:54ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_3&diff=1490972021 AIME I Problems/Problem 32021-03-11T23:20:36Z<p>Hansenhe: /* Solution */</p>
<hr />
<div>==Problem==<br />
Find the number of positive integers less than <math>1000</math> that can be expressed as the difference of two integral powers of <math>2.</math><br />
<br />
==Solution==<br />
<math>\binom{11}{2} = 55</math> We need to subtract 5 since <math>2^10 - 2^0, 2^1, 2^2, 2^3, 2^4</math> don't work. <math>55-5=\boxed{050}</math> ~hansenhe<br />
<br />
==See also==<br />
{{AIME box|year=2021|n=I|num-b=2|num-a=4}}<br />
<br />
[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Hansenhehttps://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_3&diff=1490962021 AIME I Problems/Problem 32021-03-11T23:20:00Z<p>Hansenhe: /* Solution */</p>
<hr />
<div>==Problem==<br />
Find the number of positive integers less than <math>1000</math> that can be expressed as the difference of two integral powers of <math>2.</math><br />
<br />
==Solution==<br />
<math>\binom{11}{2} = 55 \implies 55 - 5 = 50</math>. We need to subtract 5 since <math>2^10 - 2^0, 2^1, 2^2, 2^3, 2^4</math> don't work. ~hansenhe<br />
<br />
==See also==<br />
{{AIME box|year=2021|n=I|num-b=2|num-a=4}}<br />
<br />
[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Hansenhehttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_19&diff=1489662010 AMC 10B Problems/Problem 192021-03-09T03:51:38Z<p>Hansenhe: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
<br />
A circle with center <math>O</math> has area <math>156\pi</math>. Triangle <math>ABC</math> is equilateral, <math>\overline{BC}</math> is a chord on the circle, <math>OA = 4\sqrt{3}</math>, and point <math>O</math> is outside <math>\triangle ABC</math>. What is the side length of <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A)}\ 2\sqrt{3} \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 4\sqrt{3} \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18</math><br />
<br />
==Solution 1==<br />
The formula for the area of a circle is <math>\pi r^2</math> so the radius of this circle is <math>\sqrt{156}.</math><br />
<br />
Because <math>OA=4\sqrt{3} < \sqrt{156}, A</math> must be in the interior of circle <math>O.</math><br />
<br />
<center><asy><br />
unitsize(3mm);<br />
defaultpen(linewidth(.8pt)+fontsize(11pt));<br />
dotfactor=3;<br />
<br />
real r=sqrt(156);<br />
pair A=(0,sqrt(48)), B=(-3,sqrt(147)), C=(3,sqrt(147));<br />
pair O=(0,0);<br />
pair X=(0,7sqrt(3));<br />
path outer=Circle(O,r);<br />
draw(outer);<br />
draw(A--B--C--cycle);<br />
draw(O--X); draw(O--B);<br />
<br />
pair[] ps={A,B,C,O,X};<br />
dot(ps);<br />
<br />
label("$A$",A,SE);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$O$",O,S);<br />
label("$X$",X,N);<br />
label("$s$",A--C,SE);<br />
label("$\frac{s}{2}$",B--X,N);<br />
label("$\frac{s\sqrt{3}}{2}$",A--X,NE);<br />
label("$\sqrt{156}$",O--B,SW);<br />
label("$4\sqrt{3}$",A--O,E);<br />
</asy></center><br />
<br />
Let <math>s</math> be the unknown value, the sidelength of the triangle, and let <math>X</math> be the point on <math>BC</math> where <math>OX \perp BC.</math> Since <math>\triangle ABC</math> is equilateral, <math>BX=\frac{s}{2}</math> and <math>AX=\frac{s\sqrt{3}}{2}.</math> We are given <math>AO=4\sqrt{3}.</math> Use the [[Pythagorean Theorem]] and solve for <math>s.</math><br />
<br />
<cmath>\begin{align*}<br />
(\sqrt{156})^2 &= \left(\frac{s}{2}\right)^2 + \left( \frac{s\sqrt{3}}{2} + 4\sqrt{3} \right)^2\\<br />
156 &= \frac14s^2 + \frac34s^2 + 12s + 48\\<br />
0 &= s^2 + 12s - 108\\<br />
0 &= (s-6)(s+18)\\<br />
s &= \boxed{\textbf{(B)}\ 6}<br />
\end{align*} </cmath><br />
<br />
==Solution 2==<br />
We can use the same diagram as Solution 1 and label the side length of <math>\triangle ABC</math> as <math>s</math>. Using congruent triangles, namely the two triangles <math>\triangle BOA</math> and <math>\triangle COA</math>, we get that <math>\angle BAO = \angle CAO \implies \angle BAO = \frac{360-60}{2} = 150</math>. From this, we can use the [[Law of Cosines]], to get <cmath>s^2 + (4 \sqrt{3})^2 - 2 \times s \times 4 \sqrt{3} \times - \frac{\sqrt{3}}{2} = (2 \sqrt{39})^2</cmath> Simplifying, we get <cmath>s^2 + 12s + 48 = 156 \implies s^2 + 12s - 108 = 0</cmath> We can factor this to get <cmath>(x-6)(x+18)</cmath> Obviously, we want the positive solution to get <math>\boxed{\textbf{(B)}\ 6}</math><br />
~hansenhe<br />
<br />
==Video Solution==<br />
https://youtu.be/FQO-0E2zUVI?t=906<br />
<br />
~IceMatrix<br />
<br />
==See Also==<br />
{{AMC10 box|year=2010|ab=B|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Hansenhehttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_19&diff=1489652010 AMC 10B Problems/Problem 192021-03-09T03:50:51Z<p>Hansenhe: /* Solution 2 */</p>
<hr />
<div>== Problem ==<br />
<br />
A circle with center <math>O</math> has area <math>156\pi</math>. Triangle <math>ABC</math> is equilateral, <math>\overline{BC}</math> is a chord on the circle, <math>OA = 4\sqrt{3}</math>, and point <math>O</math> is outside <math>\triangle ABC</math>. What is the side length of <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A)}\ 2\sqrt{3} \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 4\sqrt{3} \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18</math><br />
<br />
==Solution 1==<br />
The formula for the area of a circle is <math>\pi r^2</math> so the radius of this circle is <math>\sqrt{156}.</math><br />
<br />
Because <math>OA=4\sqrt{3} < \sqrt{156}, A</math> must be in the interior of circle <math>O.</math><br />
<br />
<center><asy><br />
unitsize(3mm);<br />
defaultpen(linewidth(.8pt)+fontsize(11pt));<br />
dotfactor=3;<br />
<br />
real r=sqrt(156);<br />
pair A=(0,sqrt(48)), B=(-3,sqrt(147)), C=(3,sqrt(147));<br />
pair O=(0,0);<br />
pair X=(0,7sqrt(3));<br />
path outer=Circle(O,r);<br />
draw(outer);<br />
draw(A--B--C--cycle);<br />
draw(O--X); draw(O--B);<br />
<br />
pair[] ps={A,B,C,O,X};<br />
dot(ps);<br />
<br />
label("$A$",A,SE);<br />
label("$B$",B,NW);<br />
label("$C$",C,NE);<br />
label("$O$",O,S);<br />
label("$X$",X,N);<br />
label("$s$",A--C,SE);<br />
label("$\frac{s}{2}$",B--X,N);<br />
label("$\frac{s\sqrt{3}}{2}$",A--X,NE);<br />
label("$\sqrt{156}$",O--B,SW);<br />
label("$4\sqrt{3}$",A--O,E);<br />
</asy></center><br />
<br />
Let <math>s</math> be the unknown value, the sidelength of the triangle, and let <math>X</math> be the point on <math>BC</math> where <math>OX \perp BC.</math> Since <math>\triangle ABC</math> is equilateral, <math>BX=\frac{s}{2}</math> and <math>AX=\frac{s\sqrt{3}}{2}.</math> We are given <math>AO=4\sqrt{3}.</math> Use the [[Pythagorean Theorem]] and solve for <math>s.</math><br />
<br />
<cmath>\begin{align*}<br />
(\sqrt{156})^2 &= \left(\frac{s}{2}\right)^2 + \left( \frac{s\sqrt{3}}{2} + 4\sqrt{3} \right)^2\\<br />
156 &= \frac14s^2 + \frac34s^2 + 12s + 48\\<br />
0 &= s^2 + 12s - 108\\<br />
0 &= (s-6)(s+18)\\<br />
s &= \boxed{\textbf{(B)}\ 6}<br />
\end{align*} </cmath><br />
<br />
==Solution 2==<br />
We can use the same diagram as Solution 1 and label the side length of <math>\triangle ABC</math> as <math>s</math>. Using congruent triangles, namely the two triangles <math>\triangle BOA</math> and <math>\triangle COA</math>, we get that <math>\angle BAO = \angle CAO \implies \angle BAO = \frac{360-60}{2} = 150</math>. From this, we can use the Law of Cosines, to get <cmath>s^2 + (4 \sqrt{3})^2 - 2 \times s \times 4 \sqrt{3} \times - \frac{\sqrt{3}}{2} = (2 \sqrt{39})^2</cmath> Simplifying, we get <cmath>s^2 + 12s + 48 = 156 \implies s^2 + 12s - 108 = 0</cmath> We can factor this to get <cmath>(x-6)(x+18)</cmath> Obviously, we want the positive solution to get <math>\boxed{\textbf{(B)}\ 6}</math><br />
~hansenhe<br />
<br />
==Video Solution==<br />
https://youtu.be/FQO-0E2zUVI?t=906<br />
<br />
~IceMatrix<br />
<br />
==See Also==<br />
{{AMC10 box|year=2010|ab=B|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Hansenhehttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_13&diff=1475812020 AMC 10A Problems/Problem 132021-02-20T19:49:41Z<p>Hansenhe: incorrect</p>
<hr />
<div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #11]] and [[2020 AMC 10A Problems|2020 AMC 10A #13]]}}<br />
<br />
==Problem==<br />
<br />
A frog sitting at the point <math>(1, 2)</math> begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length <math>1</math>, and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices <math>(0,0), (0,4), (4,4),</math> and <math>(4,0)</math>. What is the probability that the sequence of jumps ends on a vertical side of the square<math>?</math><br />
<br />
<math> \textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78 </math><br />
<br />
==Solution 1==<br />
Drawing out the square, it's easy to see that if the frog goes to the left, it will immediately hit a vertical end of the square. Therefore, the probability of this happening is <math>\frac{1}{4} \cdot 1 = \frac{1}{4}</math>. If the frog goes to the right, it will be in the center of the square at <math>(2,2)</math>, and by symmetry (since the frog is equidistant from all sides of the square), the chance it will hit a vertical side of a square is <math>\frac{1}{2}</math>. The probability of this happening is <math>\frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}</math>.<br />
<br />
<br />
If the frog goes either up or down, it will hit a line of symmetry along the corner it is closest to and furthest to, and again, is equidistant relating to the two closer sides and also equidistant relating the two further sides. The probability for it to hit a vertical wall is <math>\frac{1}{2}</math>. Because there's a <math>\frac{1}{2}</math> chance of the frog going up and down, the total probability for this case is <math>\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}</math> and summing up all the cases, <math>\frac{1}{4} + \frac{1}{8} + \frac{1}{4} = \frac{5}{8} \implies \boxed{\textbf{(B) } \frac{5}{8}}</math>.<br />
<br />
==Solution 2==<br />
Let's say we have our four by four grid and we work this out by casework. A is where the frog is, while B and C are possible locations for his second jump, while O is everything else. If we land on a C, we have reached the vertical side. However, if we land on a B, we can see that there is an equal chance of reaching the horizontal or vertical side, since we are symmetrically between them. So we have the probability of landing on a C is 1/4, while B is 3/4. Since C means that we have "succeeded", while B means that we have a half chance, we compute <math>1 \cdot C + \frac{1}{2} \cdot B</math>. <br />
<br />
<br />
<cmath>1 \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{3}{4}</cmath><br />
<cmath>\frac{1}{4} + \frac{3}{8}</cmath><br />
We get <math>\frac{5}{8}</math>, or <math>B</math><br />
<cmath>\text{O O O O O}</cmath> <br />
<cmath>\text{O B O O O}</cmath><br />
<cmath>\text{C A B O O}</cmath><br />
<cmath>\text{O B O O O}</cmath><br />
<cmath>\text{O O O O O}</cmath><br />
-yeskay<br />
<br />
==Solution 3==<br />
If the frog is on one of the 2 diagonals, the chance of landing on vertical or horizontal each becomes <math>\frac{1}{2}</math>. Since it starts on <math>(1,2)</math>, there is a <math>\frac{3}{4}</math> chance (up, down, or right) it will reach a diagonal on the first jump and <math>\frac{1}{4}</math> chance (left) it will reach the vertical side. The probablity of landing on a vertical is <math>\frac{1}{4}+\frac{3}{4} \cdot \frac{1}{2}=\boxed{\textbf{(B)} \frac{5}{8}}</math>.<br />
- Lingjun<br />
<br />
==Solution 4 (Complete States)==<br />
Let <math>P_{(x,y)}</math> denote the probability of the frog's sequence of jumps ends with it hitting a vertical edge when it is at <math>(x,y)</math>. Note that <math>P_{(1,2)}=P_{(3,2)}</math> by reflective symmetry over the line <math>x=2</math>. Similarly, <math>P_{(1,1)}=P_{(1,3)}=P_{(3,1)}=P_{(3,3)}</math>, and <math>P_{(2,1)}=P_{(2,3)}</math>. <br />
Now we create equations for the probabilities at each of these points/states by considering the probability of going either up, down, left, or right from that point: <br />
<cmath>P_{(1,2)}=\frac{1}{4}+\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}</cmath><br />
<cmath>P_{(2,2)}=\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}</cmath><br />
<cmath>P_{(1,1)}=\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}</cmath><br />
<cmath>P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}</cmath><br />
We have a system of <math>4</math> equations in <math>4</math> variables, so we can solve for each of these probabilities. Plugging the second equation into the fourth equation gives <br />
<cmath>P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}\left(\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\right)</cmath><br />
<cmath>P_{(2,1)}=\frac{8}{7}\left(\frac{1}{2}P_{(1,1)}+\frac{1}{8}P_{(1,2)}\right)=\frac{4}{7}P_{(1,1)}+\frac{1}{7}P_{(1,2)}</cmath><br />
Plugging in the third equation into this gives <br />
<cmath>P_{(2,1)}=\frac{4}{7}\left(\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\right)+\frac{1}{7}P_{(1,2)}</cmath><br />
<cmath>P_{(2,1)}=\frac{7}{6}\left(\frac{1}{7}+\frac{2}{7}P_{(1,2)}\right)=\frac{1}{6}+\frac{1}{3}P_{(1,2)}\text{ (*)}</cmath><br />
Next, plugging in the second and third equation into the first equation yields <br />
<cmath>P_{(1,2)}=\frac{1}{4}+\frac{1}{2}\left(\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\right)+\frac{1}{4}\left(\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\right)</cmath><br />
<cmath>P_{(1,2)}=\frac{3}{8}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}</cmath><br />
Now plugging in (*) into this, we get <br />
<cmath>P_{(1,2)}=\frac{3}{8}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}\left(\frac{1}{6}+\frac{1}{3}P_{(1,2)}\right)</cmath><br />
<cmath>P_{(1,2)}=\frac{3}{2}\cdot\frac{5}{12}=\boxed{\textbf{(B) }\frac{5}{8}}</cmath><br />
-mathisawesome2169<br />
<br />
==Video Solution 1==<br />
[https://www.youtube.com/watch?v=ZGwAasE32Y4&t=280s IceMatrix's Solution (Starts at 4:40)]<br />
<br />
==Video Solution 2==<br />
<br />
https://youtu.be/qNaN0BlIsw0<br />
<br />
==Video Solution 3==<br />
On The Spot STEM<br />
<br />
https://youtu.be/xGs7BjQbGYU<br />
<br />
==Video Solution 4==<br />
https://youtu.be/0m4lbXSUV1I<br />
<br />
~savannahsolver<br />
<br />
== Video Solution 5 ==<br />
https://youtu.be/IRyWOZQMTV8?t=5173<br />
<br />
~ pi_is_3.14<br />
<br />
== Video Solution 6 ==<br />
https://www.youtube.com/watch?v=R220vbM_my8<br />
<br />
~ amritvignesh0719062.0<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2020|ab=A|num-b=10|num-a=12}}<br />
{{AMC10 box|year=2020|ab=A|num-b=12|num-a=14}}<br />
<br />
[[Category:Introductory Probability Problems]]<br />
{{MAA Notice}}</div>Hansenhehttps://artofproblemsolving.com/wiki/index.php?title=Incircle&diff=147319Incircle2021-02-17T23:12:20Z<p>Hansenhe: </p>
<hr />
<div>{{stub}}<br />
<br />
<br />
An '''incircle''' of a [[convex]] [[polygon]] is a [[circle]] which is inside the figure and [[tangent line | tangent]] to each side. Every [[triangle]] and [[regular polygon]] has a unique incircle, but in general polygons with 4 or more sides (such as non-[[square (geometry) | square]] [[rectangle]]s) do not have an incircle. A quadrilateral that does have an incircle is called a [[Tangential Quadrilateral]]. For a triangle, the center of the incircle is the [[Incenter]].<br />
<br />
{{stub}}<br />
<br />
<br />
An '''incircle''' of a [[convex]] [[polygon]] is a [[circle]] which is inside the figure and [[tangent line | tangent]] to each side. Every [[triangle]] and [[regular polygon]] has a unique incircle, but in general polygons with 4 or more sides (such as non-[[square (geometry) | square]] [[rectangle]]s) do not have an incircle. A quadrilateral that does have an incircle is called a [[Tangential Quadrilateral]]. For a triangle, the center of the incircle is the [[Incenter]], where the [[incircle]] is the largest circle that can be inscribed in the polygon. The [[Incenter]] can be constructed by drawing the intersection of angle bisectors.<br />
<br />
==Formulas==<br />
*The radius of an incircle of a triangle (the inradius) with sides <math>a,b,c</math> and area <math>A</math> is <math>r =</math> <math>\frac{2A}{a+b+c}.</math><br />
*The [[area]] of any [[triangle]] is <math>r * s,</math> where <math>s</math> is the [[Semiperimeter]] of the [[triangle]].<br />
*The formula above can be simplified with Heron's Formula, yielding <math>r = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}.</math><br />
*The [[radius]] of an incircle of a right triangle (the inradius) with legs <math>a,b</math> and hypotenuse <math>c</math> is <math>r=\frac{ab}{a+b+c}=\frac{a+b-c}{2}</math>.<br />
*For any polygon with an incircle, <math>A=sr</math>, where <math>A</math> is the area, <math>s</math> is the semi perimeter, and <math>r</math> is the inradius.<br />
*The coordinates of the incenter (center of incircle) are <math>(\dfrac{aA_x+bB_x+cC_x}{a+b+c}, \dfrac{aA_y+bB_y+cC_y}{a+b+c})</math>, if the coordinates of each vertex are <math>A(A_x, A_y)</math>, <math>B(B_x, B_y)</math>, and <math>C(C_x, C_y)</math>, the side opposite of <math>A</math> has length <math>a</math>, the side opposite of <math>B</math> has length <math>b</math>, and the side opposite of <math>C</math> has length <math>c</math>.<br />
<br />
*The formula for the [[semiperimeter]] is <math>s=\frac{a+b+c}{2}</math>.<br />
<br />
*The [[area]] of the [[triangle]] by [[Heron's Formula]] is <math>A=\sqrt{s(s-a)(s-b)(s-c)}</math>.<br />
<br />
==See also==<br />
*[[Circumradius]]<br />
*[[Inradius]]<br />
*[[Kimberling center]]<br />
<br />
[[Category:Geometry]]<br />
Click here to learn about the orthocenter, and Line's Tangent</div>Hansenhehttps://artofproblemsolving.com/wiki/index.php?title=Mass_points&diff=146515Mass points2021-02-13T23:24:26Z<p>Hansenhe: more formatting</p>
<hr />
<div>'''Mass points''' is a technique in [[Euclidean geometry]] that can greatly simplify the proofs of many theorems concerning [[polygon]]s, and is helpful in solving complex geometry problems involving lengths. In essence, it involves using a local [[coordinate system]] to identify [[point]]s by the [[ratio]]s into which they divide [[line segment]]s. Mass points are generalized by [[barycentric coordinates]].<br />
<br />
Mass point geometry was invented by Franz Mobius in 1827 along with his theory of homogeneous coordinates. The technique did not catch on until the 1960s when New York high school students made it popular. The technique greatly simplifies certain problems. Mass point geometry involves systematically assigning 'weights' to points using ratios of lengths relating vertices, which can then be used to deduce other lengths, using the fact that the lengths must be inversely proportional to their weight (just like a balanced lever). Additionally, the point dividing the line has a mass equal to the sum of the weights on either end of the line (like the fulcrum of a lever).<br />
<br />
The way to systematically assign weights to the points involves first choosing a point for the entire figure to balance around. From there, WLOG a first weight can be assigned. From the first weight, others can be derived using a few simple rules. Any line passing this central point will balance the figure. If two points balance, the product of the mass and distance from a line of balance of one point will equal the product of the mass and distance from the same line of balance of the other point. If two points are balanced, the point on the balancing line used to balance them has a mass of the sum of the masses of the two points.<br />
<br />
== Examples ==<br />
== Example 1==<br />
Consider a triangle <math>ABC</math> with its three [[Median_(geometry)|median]]s drawn, with the intersection points being <math>D, E, F,</math> corresponding to <math>AB, BC,</math> and <math>AC</math> respectively. Thus, if we label point <math>A</math> with a weight of <math>1</math>, <math>B</math> must also have a weight of <math>1</math> since <math>A</math> and <math>B</math> are equidistant from <math>D</math>. By the same process, we find <math>C</math> must also have a weight of 1. Now, since <math>A</math> and <math>B</math> both have a weight of <math>1</math>, <math>D</math> must have a weight of <math>2</math> (as is true for <math>E</math> and <math>F</math>). Thus, if we label the centroid <math>P</math>, we can deduce that <math>DP:PC</math> is <math>1:2</math> - the inverse ratio of their weights.<br />
== Example 2==<br />
<math>\triangle ABC</math> has point <math>D</math> on <math>AB</math>, point <math>E</math> on <math>BC</math>, and point <math>F</math> on <math>AC</math>. <math>AE</math>, <math>CD</math>, and <math>BF</math> intersect at point <math>G</math>. The ratio <math>AD:DB</math> is <math>3:5</math> and the ratio <math>CE:EB</math> is <math>8:3</math>. Find the ratio of <math>FG:GB</math><br />
<br />
==Problems==<br />
[[2019 AMC 8 Problems/Problem 24]]<br />
<br />
[[2016 AMC 10A Problems/Problem 19]]<br />
<br />
[[2013 AMC 10B Problems/Problem 16]]<br />
<br />
[[2004 AMC 10B Problems/Problem 20]]<br />
<br />
[[2016 AMC 12A Problems/Problem 12]]<br />
<br />
[[2009 AIME I Problems/Problem 5]]<br />
<br />
[[2009 AIME I Problems/Problem 4]]<br />
<br />
[[2001 AIME I Problems/Problem 7 ]]<br />
<br />
[[2011 AIME II Problems/Problem 4]]<br />
<br />
[[ 1992 AIME Problems/Problem 14 ]]<br />
<br />
[[ 1988 AIME Problems/Problem 12]]<br />
<br />
[[1989 AIME Problems/Problem 15]]<br />
<br />
[[1985 AIME Problems/Problem 6]]<br />
<!-- Previous external links led to errors, removed--><br />
<br />
==Video Lecture==<br />
<br />
The Central NC Math Group recently released a lecture on Mass Points and Barycentric Coordinates, which you can view at https://www.youtube.com/watch?v=KQim7-wrwL0.<br />
<br />
<br />
<br />
[[Category:Definition]]<br />
[[Category:Geometry]]</div>Hansenhehttps://artofproblemsolving.com/wiki/index.php?title=Mass_points&diff=146514Mass points2021-02-13T23:23:45Z<p>Hansenhe: formatting</p>
<hr />
<div>'''Mass points''' is a technique in [[Euclidean geometry]] that can greatly simplify the proofs of many theorems concerning [[polygon]]s, and is helpful in solving complex geometry problems involving lengths. In essence, it involves using a local [[coordinate system]] to identify [[point]]s by the [[ratio]]s into which they divide [[line segment]]s. Mass points are generalized by [[barycentric coordinates]].<br />
<br />
Mass point geometry was invented by Franz Mobius in 1827 along with his theory of homogeneous coordinates. The technique did not catch on until the 1960s when New York high school students made it popular. The technique greatly simplifies certain problems. Mass point geometry involves systematically assigning 'weights' to points using ratios of lengths relating vertices, which can then be used to deduce other lengths, using the fact that the lengths must be inversely proportional to their weight (just like a balanced lever). Additionally, the point dividing the line has a mass equal to the sum of the weights on either end of the line (like the fulcrum of a lever).<br />
<br />
The way to systematically assign weights to the points involves first choosing a point for the entire figure to balance around. From there, WLOG a first weight can be assigned. From the first weight, others can be derived using a few simple rules. Any line passing this central point will balance the figure. If two points balance, the product of the mass and distance from a line of balance of one point will equal the product of the mass and distance from the same line of balance of the other point. If two points are balanced, the point on the balancing line used to balance them has a mass of the sum of the masses of the two points.<br />
<br />
== Examples ==<br />
==1==<br />
Consider a triangle <math>ABC</math> with its three [[Median_(geometry)|median]]s drawn, with the intersection points being <math>D, E, F,</math> corresponding to <math>AB, BC,</math> and <math>AC</math> respectively. Thus, if we label point <math>A</math> with a weight of <math>1</math>, <math>B</math> must also have a weight of <math>1</math> since <math>A</math> and <math>B</math> are equidistant from <math>D</math>. By the same process, we find <math>C</math> must also have a weight of 1. Now, since <math>A</math> and <math>B</math> both have a weight of <math>1</math>, <math>D</math> must have a weight of <math>2</math> (as is true for <math>E</math> and <math>F</math>). Thus, if we label the centroid <math>P</math>, we can deduce that <math>DP:PC</math> is <math>1:2</math> - the inverse ratio of their weights.<br />
==2==<br />
<math>\triangle ABC</math> has point <math>D</math> on <math>AB</math>, point <math>E</math> on <math>BC</math>, and point <math>F</math> on <math>AC</math>. <math>AE</math>, <math>CD</math>, and <math>BF</math> intersect at point <math>G</math>. The ratio <math>AD:DB</math> is <math>3:5</math> and the ratio <math>CE:EB</math> is <math>8:3</math>. Find the ratio of <math>FG:GB</math><br />
<br />
==Problems==<br />
[[2019 AMC 8 Problems/Problem 24]]<br />
<br />
[[2016 AMC 10A Problems/Problem 19]]<br />
<br />
[[2013 AMC 10B Problems/Problem 16]]<br />
<br />
[[2004 AMC 10B Problems/Problem 20]]<br />
<br />
[[2016 AMC 12A Problems/Problem 12]]<br />
<br />
[[2009 AIME I Problems/Problem 5]]<br />
<br />
[[2009 AIME I Problems/Problem 4]]<br />
<br />
[[2001 AIME I Problems/Problem 7 ]]<br />
<br />
[[2011 AIME II Problems/Problem 4]]<br />
<br />
[[ 1992 AIME Problems/Problem 14 ]]<br />
<br />
[[ 1988 AIME Problems/Problem 12]]<br />
<br />
[[1989 AIME Problems/Problem 15]]<br />
<br />
[[1985 AIME Problems/Problem 6]]<br />
<!-- Previous external links led to errors, removed--><br />
<br />
==Video Lecture==<br />
<br />
The Central NC Math Group recently released a lecture on Mass Points and Barycentric Coordinates, which you can view at https://www.youtube.com/watch?v=KQim7-wrwL0.<br />
<br />
<br />
<br />
[[Category:Definition]]<br />
[[Category:Geometry]]</div>Hansenhehttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_20&diff=1445822018 AMC 10A Problems/Problem 202021-02-02T04:04:13Z<p>Hansenhe: </p>
<hr />
<div>{{duplicate|[[2018 AMC 12A Problems|2018 AMC 12A #15]] and [[2018 AMC 10A Problems|2018 AMC 10A #20]]}}<br />
<br />
==Problem==<br />
<br />
A scanning code consists of a <math>7 \times 7</math> grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of <math>49</math> squares. A scanning code is called <math>\textit{symmetric}</math> if its look does not change when the entire square is rotated by a multiple of <math>90 ^{\circ}</math> counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides. What is the total number of possible symmetric scanning codes?<br />
<br />
<math>\textbf{(A)} \text{ 510} \qquad \textbf{(B)} \text{ 1022} \qquad \textbf{(C)} \text{ 8190} \qquad \textbf{(D)} \text{ 8192} \qquad \textbf{(E)} \text{ 65,534}</math><br />
==Solution 1==<br />
<br />
Draw a <math>7 \times 7</math> square.<br />
<br />
<math> \begin{tabular}{|c|c|c|c|c|c|c|}<br />
\hline<br />
K & J & H & G & H & J & K \\<br />
\hline<br />
J & F & E & D & E & F & J \\<br />
\hline<br />
H & E & C & B & C & E & H \\<br />
\hline<br />
G & D & B & A & B & D & G \\<br />
\hline<br />
H & E & C & B & C & E & H \\<br />
\hline<br />
J & F & E & D & E & F & J \\<br />
\hline<br />
K & J & H & G & H & J & K \\<br />
\hline<br />
\end{tabular} </math><br />
<br />
Start from the center and label all protruding cells symmetrically. (Note that "I" is left out of this labelling, so there are only 10 labels, not 11, as ending in K would suggest!)<br />
<br />
More specifically, since there are <math>4</math> given lines of symmetry (<math>2</math> diagonals, <math>1</math> vertical, <math>1</math> horizontal) and they split the plot into <math>8</math> equivalent sections, we can take just one-eighth and study it in particular. Each of these sections has <math>10</math> distinct sub-squares, whether partially or in full. So since each can be colored either white or black, we choose <math>2^{10}=1024</math> but then subtract the <math>2</math> cases where all are white or all are black. That leaves us with <math>\fbox{\textbf{(B)} \text{ 1022}}</math>.<br />
<br />
There are only ten squares we get to actually choose, and two independent choices for each, for a total of <math>2^{10} = 1024</math> codes. Two codes must be subtracted (due to the rule that there must be at least one square of each color) for an answer of <math>\fbox{\textbf{(B) }1022}</math>.<br />
<br />
<br />
Note that this problem is very similar to the 1996 AIME Problem 7 https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_7.<br />
<br />
==Solution 2==<br />
<br />
<asy><br />
size(100pt);<br />
draw((1,0)--(8,0),linewidth(0.5));<br />
draw((1,2)--(6,2),linewidth(0.5));<br />
draw((1,4)--(4,4),linewidth(0.5));<br />
draw((1,6)--(2,6),linewidth(0.5));<br />
draw((2,6)--(2,0),linewidth(0.5));<br />
draw((4,4)--(4,0),linewidth(0.5));<br />
draw((6,2)--(6,0),linewidth(0.5));<br />
draw((1,0)--(1,7),dashed+linewidth(0.5));<br />
draw((1,7)--(8,0),dashed+linewidth(0.5));<br />
</asy><br />
<br />
Imagine folding the scanning code along its lines of symmetry. There will be <math>10</math> regions which you have control over coloring. Since we must subtract off <math>2</math> cases for the all-black and all-white cases, the answer is <math>2^{10}-2=\boxed{\textbf{(B) } 1022.}</math><br />
<br />
==Solution 3==<br />
This <math>7 \times 7</math> square drawn in Solution 1 satisfies the conditions given in the problem. Calculating the number of ways of coloring it will solve the problem.<br />
<br />
<math> \begin{tabular}{|c|c|c|c|c|c|c|}<br />
\hline<br />
K & J & H & G & H & J & K \\<br />
\hline<br />
J & F & E & D & E & F & J \\<br />
\hline<br />
H & E & C & B & C & E & H \\<br />
\hline<br />
G & D & B & A & B & D & G \\<br />
\hline<br />
H & E & C & B & C & E & H \\<br />
\hline<br />
J & F & E & D & E & F & J \\<br />
\hline<br />
K & J & H & G & H & J & K \\<br />
\hline<br />
\end{tabular} </math><br />
<br />
In the grid, 10 letters are used: <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, <math>E</math>, <math>F</math>, <math>G</math>, <math>H</math>, <math>J</math>, and <math>K</math>. Each of the letters must have its own color, either white or black. This means, for example, all <math>K</math>'s must have the same color for the grid to be symmetrical.<br />
<br />
So there are <math>2^{10}</math> ways to color the grid, including a completely black grid and a completely white grid. Since the grid must contain at least one square with each color, the number of ways is <math>2^{10}-2=1024-2=</math> <math>\boxed{\textbf{(B) } 1022}</math>.<br />
<br />
==Solution 4==<br />
<br />
<math> \begin{tabular}{|c|c|c|c|c|c|c|}<br />
\hline<br />
T & T & T & X & T & T & T \\<br />
\hline<br />
T & T & T & Y& T & T & T \\<br />
\hline<br />
T & T & T & Z & T & T & T \\<br />
\hline<br />
X & Y & Z & W & Z & Y & X \\<br />
\hline<br />
T & T & T & Z & T & T & T \\<br />
\hline<br />
T & T & T & Y & T & T & T \\<br />
\hline<br />
T & T & T & X & T & T & T \\<br />
\hline<br />
\end{tabular} </math><br />
<br />
There are <math>3 \times 3</math> squares in the corners of this <math>7 \times 7</math> square, and there is a horizontal and vertical stripe through the middle. Because we need to have symmetry when the diagonals and midpoints of the large square is connected, we can create a table like this: (Different letters represent different color choices between black and white)<br />
<br />
<math> \begin{tabular}{|c|c|c|}<br />
\hline<br />
A & B & C \\<br />
\hline<br />
B & D & E \\<br />
\hline<br />
C & E & F \\<br />
\hline<br />
\end{tabular}</math><br />
(Note that coloring one <math>3 \times 3</math> square will also determine the colorings of the other <math>3</math> because the large <math>7 \times 7</math> square must look the same when it is rotated by <math>90^\circ</math>)<br />
<br />
There are <math>6</math> different letters and <math>2</math> choices of color (black and white) for each letter, so there are <math>2^6=64</math> colorings of a proper <math>3 \times 3</math> square. Now, all that's left are the horizontal and vertical stripes. Using similar logic, we can see that there are <math>4</math> different letters, so there are <math>2^4=16</math> different colorings. Multiplying them together gives <math>16 \times 64 = 1024</math>. Going back to the question, we see that "there must be at least one square of each color in this grid of <math>49</math> squares." We must then eliminate <math>2</math> options: an all-black grid and an all-white grid. <math>1024-2= \boxed{\textbf{(B)} 1022} \\ \phantom{}</math> <br />
-hansenhe<br />
<br />
==Video Solution==<br />
https://youtu.be/M22S82Am2zM<br />
<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2018|ab=A|num-b=19|num-a=21}}<br />
{{AMC12 box|year=2018|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]</div>Hansenhehttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_20&diff=1445812018 AMC 10A Problems/Problem 202021-02-02T03:50:29Z<p>Hansenhe: Add new solution</p>
<hr />
<div>{{duplicate|[[2018 AMC 12A Problems|2018 AMC 12A #15]] and [[2018 AMC 10A Problems|2018 AMC 10A #20]]}}<br />
<br />
==Problem==<br />
<br />
A scanning code consists of a <math>7 \times 7</math> grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of <math>49</math> squares. A scanning code is called <math>\textit{symmetric}</math> if its look does not change when the entire square is rotated by a multiple of <math>90 ^{\circ}</math> counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides. What is the total number of possible symmetric scanning codes?<br />
<br />
<math>\textbf{(A)} \text{ 510} \qquad \textbf{(B)} \text{ 1022} \qquad \textbf{(C)} \text{ 8190} \qquad \textbf{(D)} \text{ 8192} \qquad \textbf{(E)} \text{ 65,534}</math><br />
==Solution 1==<br />
<br />
Draw a <math>7 \times 7</math> square.<br />
<br />
<math> \begin{tabular}{|c|c|c|c|c|c|c|}<br />
\hline<br />
K & J & H & G & H & J & K \\<br />
\hline<br />
J & F & E & D & E & F & J \\<br />
\hline<br />
H & E & C & B & C & E & H \\<br />
\hline<br />
G & D & B & A & B & D & G \\<br />
\hline<br />
H & E & C & B & C & E & H \\<br />
\hline<br />
J & F & E & D & E & F & J \\<br />
\hline<br />
K & J & H & G & H & J & K \\<br />
\hline<br />
\end{tabular} </math><br />
<br />
Start from the center and label all protruding cells symmetrically. (Note that "I" is left out of this labelling, so there are only 10 labels, not 11, as ending in K would suggest!)<br />
<br />
More specifically, since there are <math>4</math> given lines of symmetry (<math>2</math> diagonals, <math>1</math> vertical, <math>1</math> horizontal) and they split the plot into <math>8</math> equivalent sections, we can take just one-eighth and study it in particular. Each of these sections has <math>10</math> distinct sub-squares, whether partially or in full. So since each can be colored either white or black, we choose <math>2^{10}=1024</math> but then subtract the <math>2</math> cases where all are white or all are black. That leaves us with <math>\fbox{\textbf{(B)} \text{ 1022}}</math>.<br />
<br />
There are only ten squares we get to actually choose, and two independent choices for each, for a total of <math>2^{10} = 1024</math> codes. Two codes must be subtracted (due to the rule that there must be at least one square of each color) for an answer of <math>\fbox{\textbf{(B) }1022}</math>.<br />
<br />
<br />
Note that this problem is very similar to the 1996 AIME Problem 7 https://artofproblemsolving.com/wiki/index.php/1996_AIME_Problems/Problem_7.<br />
<br />
==Solution 2==<br />
<br />
<asy><br />
size(100pt);<br />
draw((1,0)--(8,0),linewidth(0.5));<br />
draw((1,2)--(6,2),linewidth(0.5));<br />
draw((1,4)--(4,4),linewidth(0.5));<br />
draw((1,6)--(2,6),linewidth(0.5));<br />
draw((2,6)--(2,0),linewidth(0.5));<br />
draw((4,4)--(4,0),linewidth(0.5));<br />
draw((6,2)--(6,0),linewidth(0.5));<br />
draw((1,0)--(1,7),dashed+linewidth(0.5));<br />
draw((1,7)--(8,0),dashed+linewidth(0.5));<br />
</asy><br />
<br />
Imagine folding the scanning code along its lines of symmetry. There will be <math>10</math> regions which you have control over coloring. Since we must subtract off <math>2</math> cases for the all-black and all-white cases, the answer is <math>2^{10}-2=\boxed{\textbf{(B) } 1022.}</math><br />
<br />
==Solution 3==<br />
This <math>7 \times 7</math> square drawn in Solution 1 satisfies the conditions given in the problem. Calculating the number of ways of coloring it will solve the problem.<br />
<br />
<math> \begin{tabular}{|c|c|c|c|c|c|c|}<br />
\hline<br />
K & J & H & G & H & J & K \\<br />
\hline<br />
J & F & E & D & E & F & J \\<br />
\hline<br />
H & E & C & B & C & E & H \\<br />
\hline<br />
G & D & B & A & B & D & G \\<br />
\hline<br />
H & E & C & B & C & E & H \\<br />
\hline<br />
J & F & E & D & E & F & J \\<br />
\hline<br />
K & J & H & G & H & J & K \\<br />
\hline<br />
\end{tabular} </math><br />
<br />
In the grid, 10 letters are used: <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, <math>E</math>, <math>F</math>, <math>G</math>, <math>H</math>, <math>J</math>, and <math>K</math>. Each of the letters must have its own color, either white or black. This means, for example, all <math>K</math>'s must have the same color for the grid to be symmetrical.<br />
<br />
So there are <math>2^{10}</math> ways to color the grid, including a completely black grid and a completely white grid. Since the grid must contain at least one square with each color, the number of ways is <math>2^{10}-2=1024-2=</math> <math>\boxed{\textbf{(B) } 1022}</math>.<br />
<br />
==Solution 4==<br />
<br />
<math> \begin{tabular}{|c|c|c|c|c|c|c|}<br />
\hline<br />
T & T & T & X & T & T & T \\<br />
\hline<br />
T & T & T & Y& T & T & T \\<br />
\hline<br />
T & T & T & Z & T & T & T \\<br />
\hline<br />
X & Y & Z & W & Z & Y & X \\<br />
\hline<br />
T & T & T & Z & T & T & T \\<br />
\hline<br />
T & T & T & Y & T & T & T \\<br />
\hline<br />
T & T & T & X & T & T & T \\<br />
\hline<br />
\end{tabular} </math><br />
<br />
There are <math>3 \times 3</math> squares in the corners of this <math>7 \times 7</math> square, and there is a horizontal and vertical stripe through the middle. Because we need to have symmetry when the diagonals and midpoints of the large square is connected, we can create a table like this: (Different letters represent different color choices between black and white)<br />
<br />
<math> \begin{tabular}{|c|c|c|}<br />
\hline<br />
A & B & C \\<br />
\hline<br />
B & D & E \\<br />
\hline<br />
C & E & F \\<br />
\hline<br />
\end{tabular}</math><br />
<br />
There are <math>6</math> different letters and <math>2</math> choices of color (black and white) for each letter, so there are <math>2^6=64</math> colorings of a proper <math>3 \times 3</math> square. Now, all that's left are the horizontal and vertical stripes. Using similar logic, we can see that there are <math>4</math> different letters, so there are <math>2^4=16</math> different colorings. Multiplying them together gives <math>16 \times 64 = 1024</math>. Going back to the question, we see that "there must be at least one square of each color in this grid of <math>49</math> squares." We must then eliminate <math>2</math> options: an all-black grid and an all-white grid. <math>1024-2= \boxed{\textbf{(B)} 1022}</math><br />
<br />
==Video Solution==<br />
https://youtu.be/M22S82Am2zM<br />
<br />
<br />
== See Also ==<br />
<br />
{{AMC10 box|year=2018|ab=A|num-b=19|num-a=21}}<br />
{{AMC12 box|year=2018|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]</div>Hansenhehttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_16&diff=1434762010 AMC 10B Problems/Problem 162021-01-27T23:52:58Z<p>Hansenhe: edit a small error</p>
<hr />
<div>== Problem==<br />
A square of side length <math>1</math> and a circle of radius <math>\dfrac{\sqrt{3}}{3}</math> share the same center. What is the area inside the circle, but outside the square?<br />
<br />
<math>\textbf{(A)}\ \dfrac{\pi}{3}-1 \qquad \textbf{(B)}\ \dfrac{2\pi}{9}-\dfrac{\sqrt{3}}{3} \qquad \textbf{(C)}\ \dfrac{\pi}{18} \qquad \textbf{(D)}\ \dfrac{1}{4} \qquad \textbf{(E)}\ \dfrac{2\pi}{9}</math><br />
<br />
==Solution 1==<br />
The radius of the circle is <math>\frac{\sqrt{3}}{3} = \sqrt{\frac{1}{3}}</math>. Half the diagonal of the square is <math>\frac{\sqrt{1^2+1^2}}{2} = \frac{\sqrt{2}}{2} = \sqrt{\frac12}</math>. We can see that the circle passes outside the square, but the square is NOT completely contained in the circle. Therefore the picture will look something like this:<br />
<br />
<center><asy><br />
unitsize(5cm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=3;<br />
<br />
real r=sqrt(1/3);<br />
pair O=(0,0);<br />
pair W=(0.5,0.5), X=(0.5,-0.5), Y=(-0.5,-0.5), Z=(-0.5,0.5);<br />
pair A=(-sqrt(1/12),0.5), B=(sqrt(1/12),0.5);<br />
pair V=(0,0.5);<br />
path outer=Circle(O,r);<br />
draw(outer);<br />
draw(W--X--Y--Z--cycle);<br />
draw(O--A);<br />
draw(O--B);<br />
draw(V--O);<br />
<br />
pair[] ps={A,B,V,O};<br />
dot(ps);<br />
<br />
label("$O$",O,SW);<br />
label("$\frac{\sqrt{3}}{3}$",O--B,SE);<br />
label("$A$",A,NW);<br />
label("$B$",B,NE);<br />
label("$X$",V,NW);<br />
label("$a$",B--V,S);<br />
label("$\frac12$",O--V,W);<br />
</asy></center><br />
<br />
Then we proceed to find: 4 <math>\cdot</math> (area of sector marked off by the two radii - area of the triangle with sides on the square and the two radii).<br />
<br />
First we realize that the radius perpendicular to the side of the square between the two radii marking off the sector splits <math>AB</math> in half. Let this half-length be <math>a</math>. Also note that <math>OX=\frac12</math> because it is half the sidelength of the square. Because this is a right triangle, we can use the [[Pythagorean Theorem]] to solve for <math>a.</math><br />
<br />
<cmath>a^2+\left( \frac12 \right) ^2 = \left( \frac{\sqrt{3}}{3} \right) ^2</cmath><br />
<br />
Solving, <math>a= \frac{\sqrt{3}}{6}</math> and <math>2a=\frac{\sqrt{3}}{3}</math>. Since <math>AB=AO=BO</math>, <math>\triangle AOB</math> is an equilateral triangle and the central angle is <math>60^{\circ}</math>. Therefore the sector has an area <math>\pi \left( \frac{\sqrt{3}}{3} \right) ^2 \left( \frac{60}{360} \right) = \frac{\pi}{18}</math>.<br />
<br />
Now we turn to the triangle. Since it is equilateral, we can use the formula for the [[area of an equilateral triangle]] which is<br />
<br />
<cmath>\frac{s^2\sqrt{3}}{4} = \frac{\frac13 \sqrt{3}}{4} = \frac{\sqrt{3}}{12}</cmath><br />
<br />
Putting it together, we get the answer to be <math>4\cdot\left( \frac{\pi}{18}-\frac{\sqrt{3}}{12} \right)= \boxed{\textbf{(B)}\ \frac{2\pi}{9}-\frac{\sqrt{3}}{3}}</math><br />
<br />
==Solution 2 (Answer Choices)==<br />
<br />
At once, we can eliminate <math>(C)</math>, <math>(D)</math>, and <math>(E)</math> because the answer should be <math>a \times \pi + b</math>, where <math>a</math> and <math>b</math> are real numbers (not necessarily rational) not equal to <math>0</math>. Now, all that's left is <math>\frac{\pi}{3}-1</math> and <math>\frac{2 \pi}{9} - \frac{\sqrt{3}}{3}</math>. <math>\frac{\pi}{3}-1</math> is simply the area of the circle minus the square, which is definitely wrong, so the answer is <math>\boxed{\textbf{(B)} \frac{2 \pi}{9} - \frac{\sqrt{3}}{3}}</math><br />
<br />
==Video Solution==<br />
https://youtu.be/FQO-0E2zUVI<br />
<br />
~IceMatrix<br />
<br />
==See Also==<br />
{{AMC10 box|year=2010|ab=B|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Hansenhehttps://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_16&diff=1433552010 AMC 10B Problems/Problem 162021-01-27T00:47:19Z<p>Hansenhe: </p>
<hr />
<div>== Problem==<br />
A square of side length <math>1</math> and a circle of radius <math>\dfrac{\sqrt{3}}{3}</math> share the same center. What is the area inside the circle, but outside the square?<br />
<br />
<math>\textbf{(A)}\ \dfrac{\pi}{3}-1 \qquad \textbf{(B)}\ \dfrac{2\pi}{9}-\dfrac{\sqrt{3}}{3} \qquad \textbf{(C)}\ \dfrac{\pi}{18} \qquad \textbf{(D)}\ \dfrac{1}{4} \qquad \textbf{(E)}\ \dfrac{2\pi}{9}</math><br />
<br />
==Solution 1==<br />
The radius of the circle is <math>\frac{\sqrt{3}}{3} = \sqrt{\frac{1}{3}}</math>. Half the diagonal of the square is <math>\frac{\sqrt{1^2+1^2}}{2} = \frac{\sqrt{2}}{2} = \sqrt{\frac12}</math>. We can see that the circle passes outside the square, but the square is NOT completely contained in the circle. Therefore the picture will look something like this:<br />
<br />
<center><asy><br />
unitsize(5cm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=3;<br />
<br />
real r=sqrt(1/3);<br />
pair O=(0,0);<br />
pair W=(0.5,0.5), X=(0.5,-0.5), Y=(-0.5,-0.5), Z=(-0.5,0.5);<br />
pair A=(-sqrt(1/12),0.5), B=(sqrt(1/12),0.5);<br />
pair V=(0,0.5);<br />
path outer=Circle(O,r);<br />
draw(outer);<br />
draw(W--X--Y--Z--cycle);<br />
draw(O--A);<br />
draw(O--B);<br />
draw(V--O);<br />
<br />
pair[] ps={A,B,V,O};<br />
dot(ps);<br />
<br />
label("$O$",O,SW);<br />
label("$\frac{\sqrt{3}}{3}$",O--B,SE);<br />
label("$A$",A,NW);<br />
label("$B$",B,NE);<br />
label("$X$",V,NW);<br />
label("$a$",B--V,S);<br />
label("$\frac12$",O--V,W);<br />
</asy></center><br />
<br />
Then we proceed to find: 4 <math>\cdot</math> (area of sector marked off by the two radii - area of the triangle with sides on the square and the two radii).<br />
<br />
First we realize that the radius perpendicular to the side of the square between the two radii marking off the sector splits <math>AB</math> in half. Let this half-length be <math>a</math>. Also note that <math>OX=\frac12</math> because it is half the sidelength of the square. Because this is a right triangle, we can use the [[Pythagorean Theorem]] to solve for <math>a.</math><br />
<br />
<cmath>a^2+\left( \frac12 \right) ^2 = \left( \frac{\sqrt{3}}{3} \right) ^2</cmath><br />
<br />
Solving, <math>a= \frac{\sqrt{3}}{6}</math> and <math>2a=\frac{\sqrt{3}}{3}</math>. Since <math>AB=AO=BO</math>, <math>\triangle AOB</math> is an equilateral triangle and the central angle is <math>60^{\circ}</math>. Therefore the sector has an area <math>\pi \left( \frac{\sqrt{3}}{3} \right) ^2 \left( \frac{60}{360} \right) = \frac{\pi}{18}</math>.<br />
<br />
Now we turn to the triangle. Since it is equilateral, we can use the formula for the [[area of an equilateral triangle]] which is<br />
<br />
<cmath>\frac{s^2\sqrt{3}}{4} = \frac{\frac13 \sqrt{3}}{4} = \frac{\sqrt{3}}{12}</cmath><br />
<br />
Putting it together, we get the answer to be <math>4\cdot\left( \frac{\pi}{18}-\frac{\sqrt{3}}{12} \right)= \boxed{\textbf{(B)}\ \frac{2\pi}{9}-\frac{\sqrt{3}}{3}}</math><br />
<br />
==Solution 2 (Answer Choices)==<br />
<br />
At once, we can eliminate <math>(C)</math>, <math>(D)</math>, and <math>(E)</math> because the answer should be <math>a \times \pi + b</math>, where <math>a</math> and <math>b</math> are real numbers (not necessarily rational) less than <math>1</math>. Now, all that's left is <math>\frac{\pi}{3}-1</math> and <math>\frac{2 \pi}{9} - \frac{\sqrt{3}}{3}</math>. <math>\frac{\pi}{3}-1</math> is simply the area of the circle minus the square, which is definitely wrong, so the answer is <math>\boxed{\textbf{(B)} \frac{2 \pi}{9} - \frac{\sqrt{3}}{3}}</math><br />
<br />
==Video Solution==<br />
https://youtu.be/FQO-0E2zUVI<br />
<br />
~IceMatrix<br />
<br />
==See Also==<br />
{{AMC10 box|year=2010|ab=B|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Hansenhehttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_22&diff=1423482018 AMC 10B Problems/Problem 222021-01-17T01:36:22Z<p>Hansenhe: we don't want to learn from bogus/false solutions >:(</p>
<hr />
<div>==Problem==<br />
<br />
Real numbers <math>x</math> and <math>y</math> are chosen independently and uniformly at random from the interval <math>[0,1]</math>. Which of the following numbers is closest to the probability that <math>x,y,</math> and <math>1</math> are the side lengths of an obtuse triangle?<br />
<br />
<math>\textbf{(A)} \text{ 0.21} \qquad \textbf{(B)} \text{ 0.25} \qquad \textbf{(C)} \text{ 0.29} \qquad \textbf{(D)} \text{ 0.50} \qquad \textbf{(E)} \text{ 0.79}</math><br />
<br />
== Solution 1== <br />
The Pythagorean Inequality tells us that in an obtuse triangle, <math>a^{2} + b^{2} < c^{2}</math>. The triangle inequality tells us that <math>a + b > c</math>. So, we have two inequalities:<br />
<cmath>x^2 + y^2 < 1</cmath><br />
<cmath>x + y > 1</cmath><br />
The first equation is <math>\frac14</math> of a circle with radius <math>1</math>, and the second equation is a line from <math>(0, 1)</math> to <math>(1, 0)</math>.<br />
So, the area is <math>\frac{\pi}{4} - \frac12</math> which is approximately <math>\boxed{0.29}</math>, which is <math>\boxed{C}</math><br />
<br />
latex edits - srisainandan6<br />
<br />
==Solution 2 (Trig)==<br />
<br />
Note that the obtuse angle in the triangle has to be opposite the side that is always length <math>1</math>. This is because the largest angle is always opposite the largest side, and if two sides of the triangle were <math>1</math>, the last side would have to be greater than <math>1</math> to make an obtuse triangle. Using this observation, we can set up a law of cosines where the angle is opposite <math>1</math>:<br />
<br />
<cmath>1^2=x^2+y^2-2xy\cos(\theta)</cmath><br />
<br />
where <math>x</math> and <math>y</math> are the sides that go from <math>[0,1]</math> and <math>\theta</math> is the angle opposite the side of length <math>1</math>.<br />
<br />
By isolating <math>\cos(\theta)</math>, we get:<br />
<br />
<cmath>\frac{1-x^2-y^2}{-2xy} = \cos(\theta)</cmath><br />
<br />
For <math>\theta</math> to be obtuse, <math>\cos(\theta)</math> must be negative. Therefore, <math>\frac{1-x^2-y^2}{-2xy}</math> is negative. Since <math>x</math> and <math>y</math> must be positive, <math>-2xy</math> must be negative, so we must make <math>1-x^2-y^2</math> positive. From here, we can set up the inequality<br />
<cmath>x^2+y^2<1</cmath><br />
Additionally, to satisfy the definition of a triangle, we need:<br />
<cmath>x+y>1</cmath><br />
The solution should be the overlap between the two equations in the first quadrant.<br />
<br />
By observing that <math>x^2+y^2<1</math> is the equation for a circle, the amount that is in the first quadrant is <math>\frac{\pi}{4}</math>. The line can also be seen as a chord that goes from <math>(0, 1)</math> to <math>(1, 0)</math>. By cutting off the triangle of area <math>\frac{1}{2}</math> that is not part of the overlap, we get <math>\frac{\pi}{4} - \frac{1}{2} \approx \boxed{0.29}</math>.<br />
<br />
-allenle873<br />
<br />
==Video Solution==<br />
https://www.youtube.com/watch?v=GHAMU60rI5c<br />
<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Hansenhehttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_13&diff=1406532020 AMC 10A Problems/Problem 132020-12-26T18:08:51Z<p>Hansenhe: edits</p>
<hr />
<div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #11]] and [[2020 AMC 10A Problems|2020 AMC 10A #13]]}}<br />
<br />
==Problem 13==<br />
<br />
A frog sitting at the point <math>(1, 2)</math> begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length <math>1</math>, and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices <math>(0,0), (0,4), (4,4),</math> and <math>(4,0)</math>. What is the probability that the sequence of jumps ends on a vertical side of the square<math>?</math><br />
<br />
<math> \textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78 </math><br />
<br />
==Solution 1==<br />
Drawing out the square, it's easy to see that if the frog goes to the left, it will immediately hit a vertical end of the square. Therefore, the probability of this happening is <math>\frac{1}{4} * 1 = \frac{1}{4}</math>. If the frog goes to the right, it will be in the center of the square at <math>(2,2)</math>, and by symmetry (since the frog is equidistant from all sides of the square), the chance it will hit a vertical side of a square is <math>\frac{1}{2}</math>. The probability of this happening is <math>\frac{1}{4} * \frac{1}{2} = \frac{1}{8}</math>.<br />
<br />
<br />
If the frog goes either up or down, it will hit a line of symmetry along the corner it is closest to and furthest to, and again, is equidistant relating to the two closer sides and also equidistant relating the two further sides. The probability for it to hit a vertical wall is <math>\frac{1}{2}</math>. Because there's a <math>\frac{1}{2}</math> chance of the frog going up and down, the total probability for this case is <math>\frac{1}{2} * \frac{1}{2} = \frac{1}{4}</math> and summing up all the cases, <math>\frac{1}{4} + \frac{1}{8} + \frac{1}{4} = \frac{5}{8} \implies \boxed{\textbf{(B) } \frac{5}{8}}</math>.<br />
<br />
==Solution 2==<br />
Let's say we have our four by four grid and we work this out by casework. A is where the frog is, while B and C are possible locations for his second jump, while O is everything else. If we land on a C, we have reached the vertical side. However, if we land on a B, we can see that there is an equal chance of reaching the horizontal or vertical side, since we are symmetrically between them. So we have the probability of landing on a C is 1/4, while B is 3/4. Since C means that we have "succeeded", while B means that we have a half chance, we compute <math>1 \cdot C + \frac{1}{2} \cdot B</math>. <br />
<br />
<br />
<cmath>1 \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{3}{4}</cmath><br />
<cmath>\frac{1}{4} + \frac{3}{8}</cmath><br />
We get <math>\frac{5}{8}</math>, or <math>B</math><br />
<cmath>\text{O O O O O}</cmath> <br />
<cmath>\text{O B O O O}</cmath><br />
<cmath>\text{C A B O O}</cmath><br />
<cmath>\text{O B O O O}</cmath><br />
<cmath>\text{O O O O O}</cmath><br />
-yeskay<br />
<br />
==Solution 3==<br />
If the frog is on one of the 2 diagonals, the chance of landing on vertical or horizontal each becomes <math>\frac{1}{2}</math>. Since it starts on <math>(1,2)</math>, there is a <math>\frac{3}{4}</math> chance (up, down, or right) it will reach a diagonal on the first jump and <math>\frac{1}{4}</math> chance (left) it will reach the vertical side. The probablity of landing on a vertical is <math>\frac{1}{4}+\frac{3}{4} * \frac{1}{2}=\boxed{\textbf{(B)} \frac{5}{8}}</math>.<br />
- Lingjun<br />
<br />
==Solution 4 (Complete States)==<br />
Let <math>P_{(x,y)}</math> denote the probability of the frog's sequence of jumps ends with it hitting a vertical edge when it is at <math>(x,y)</math>. Note that <math>P_{(1,2)}=P_{(3,2)}</math> by reflective symmetry over the line <math>x=2</math>. Similarly, <math>P_{(1,1)}=P_{(1,3)}=P_{(3,1)}=P_{(3,3)}</math>, and <math>P_{(2,1)}=P_{(2,3)}</math>. <br />
Now we create equations for the probabilities at each of these points/states by considering the probability of going either up, down, left, or right from that point: <br />
<cmath>P_{(1,2)}=\frac{1}{4}+\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}</cmath><br />
<cmath>P_{(2,2)}=\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}</cmath><br />
<cmath>P_{(1,1)}=\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}</cmath><br />
<cmath>P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}</cmath><br />
We have a system of <math>4</math> equations in <math>4</math> variables, so we can solve for each of these probabilities. Plugging the second equation into the fourth equation gives <br />
<cmath>P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}\left(\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\right)</cmath><br />
<cmath>P_{(2,1)}=\frac{8}{7}\left(\frac{1}{2}P_{(1,1)}+\frac{1}{8}P_{(1,2)}\right)=\frac{4}{7}P_{(1,1)}+\frac{1}{7}P_{(1,2)}</cmath><br />
Plugging in the third equation into this gives <br />
<cmath>P_{(2,1)}=\frac{4}{7}\left(\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\right)+\frac{1}{7}P_{(1,2)}</cmath><br />
<cmath>P_{(2,1)}=\frac{7}{6}\left(\frac{1}{7}+\frac{2}{7}P_{(1,2)}\right)=\frac{1}{6}+\frac{1}{3}P_{(1,2)}\text{ (*)}</cmath><br />
Next, plugging in the second and third equation into the first equation yields <br />
<cmath>P_{(1,2)}=\frac{1}{4}+\frac{1}{2}\left(\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\right)+\frac{1}{4}\left(\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\right)</cmath><br />
<cmath>P_{(1,2)}=\frac{3}{8}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}</cmath><br />
Now plugging in (*) into this, we get <br />
<cmath>P_{(1,2)}=\frac{3}{8}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}\left(\frac{1}{6}+\frac{1}{3}P_{(1,2)}\right)</cmath><br />
<cmath>P_{(1,2)}=\frac{3}{2}\cdot\frac{5}{12}=\boxed{\textbf{(B) }\frac{5}{8}}</cmath><br />
-mathisawesome2169<br />
<br />
==Solution 5 (Very fast)==<br />
<br />
We can immediately note that the probability of landing on any lattice point is equal to the probability of landing on another. With this in mind, we see that each vertical "boundary" has 5 lattice points. (Remember, the "corners" count!) There are two boundary lines, so there are <math>2 \times 5 = 10</math> lattice points on our desired vertical boundary lines. The total amount of lattice points on the <math>4 \times 4</math> boundary is <math>16</math>. Using <math>\frac{P(desired)}{P(total)}</math>, we get <math>\frac{10}{16} = \boxed{\textbf{(B) }\frac{5}{8}}</math> <cmath>\phantom{}</cmath><br />
-hansenhe<br />
<br />
==Video Solution 1==<br />
[https://www.youtube.com/watch?v=ZGwAasE32Y4&t=280s IceMatrix's Solution (Starts at 4:40)]<br />
<br />
==Video Solution 2==<br />
<br />
https://youtu.be/qNaN0BlIsw0<br />
<br />
==Video Solution 3==<br />
On The Spot STEM<br />
<br />
https://youtu.be/xGs7BjQbGYU<br />
<br />
==Video Solution 4==<br />
https://youtu.be/0m4lbXSUV1I<br />
<br />
~savannahsolver<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2020|ab=A|num-b=10|num-a=12}}<br />
{{AMC10 box|year=2020|ab=A|num-b=12|num-a=14}}<br />
<br />
[[Category:Introductory Probability Problems]]<br />
{{MAA Notice}}</div>Hansenhehttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_13&diff=1404002020 AMC 10A Problems/Problem 132020-12-23T22:25:34Z<p>Hansenhe: Add solution 5</p>
<hr />
<div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #11]] and [[2020 AMC 10A Problems|2020 AMC 10A #13]]}}<br />
<br />
==Problem 13==<br />
<br />
A frog sitting at the point <math>(1, 2)</math> begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length <math>1</math>, and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices <math>(0,0), (0,4), (4,4),</math> and <math>(4,0)</math>. What is the probability that the sequence of jumps ends on a vertical side of the square<math>?</math><br />
<br />
<math> \textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78 </math><br />
<br />
==Solution 1==<br />
Drawing out the square, it's easy to see that if the frog goes to the left, it will immediately hit a vertical end of the square. Therefore, the probability of this happening is <math>\frac{1}{4} * 1 = \frac{1}{4}</math>. If the frog goes to the right, it will be in the center of the square at <math>(2,2)</math>, and by symmetry (since the frog is equidistant from all sides of the square), the chance it will hit a vertical side of a square is <math>\frac{1}{2}</math>. The probability of this happening is <math>\frac{1}{4} * \frac{1}{2} = \frac{1}{8}</math>.<br />
<br />
<br />
If the frog goes either up or down, it will hit a line of symmetry along the corner it is closest to and furthest to, and again, is equidistant relating to the two closer sides and also equidistant relating the two further sides. The probability for it to hit a vertical wall is <math>\frac{1}{2}</math>. Because there's a <math>\frac{1}{2}</math> chance of the frog going up and down, the total probability for this case is <math>\frac{1}{2} * \frac{1}{2} = \frac{1}{4}</math> and summing up all the cases, <math>\frac{1}{4} + \frac{1}{8} + \frac{1}{4} = \frac{5}{8} \implies \boxed{\textbf{(B) } \frac{5}{8}}</math>.<br />
<br />
==Solution 2==<br />
Let's say we have our four by four grid and we work this out by casework. A is where the frog is, while B and C are possible locations for his second jump, while O is everything else. If we land on a C, we have reached the vertical side. However, if we land on a B, we can see that there is an equal chance of reaching the horizontal or vertical side, since we are symmetrically between them. So we have the probability of landing on a C is 1/4, while B is 3/4. Since C means that we have "succeeded", while B means that we have a half chance, we compute <math>1 \cdot C + \frac{1}{2} \cdot B</math>. <br />
<br />
<br />
<cmath>1 \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{3}{4}</cmath><br />
<cmath>\frac{1}{4} + \frac{3}{8}</cmath><br />
We get <math>\frac{5}{8}</math>, or <math>B</math><br />
<cmath>\text{O O O O O}</cmath> <br />
<cmath>\text{O B O O O}</cmath><br />
<cmath>\text{C A B O O}</cmath><br />
<cmath>\text{O B O O O}</cmath><br />
<cmath>\text{O O O O O}</cmath><br />
-yeskay<br />
<br />
==Solution 3==<br />
If the frog is on one of the 2 diagonals, the chance of landing on vertical or horizontal each becomes <math>\frac{1}{2}</math>. Since it starts on <math>(1,2)</math>, there is a <math>\frac{3}{4}</math> chance (up, down, or right) it will reach a diagonal on the first jump and <math>\frac{1}{4}</math> chance (left) it will reach the vertical side. The probablity of landing on a vertical is <math>\frac{1}{4}+\frac{3}{4} * \frac{1}{2}=\boxed{\textbf{(B)} \frac{5}{8}}</math>.<br />
- Lingjun<br />
<br />
==Solution 4 (Complete States)==<br />
Let <math>P_{(x,y)}</math> denote the probability of the frog's sequence of jumps ends with it hitting a vertical edge when it is at <math>(x,y)</math>. Note that <math>P_{(1,2)}=P_{(3,2)}</math> by reflective symmetry over the line <math>x=2</math>. Similarly, <math>P_{(1,1)}=P_{(1,3)}=P_{(3,1)}=P_{(3,3)}</math>, and <math>P_{(2,1)}=P_{(2,3)}</math>. <br />
Now we create equations for the probabilities at each of these points/states by considering the probability of going either up, down, left, or right from that point: <br />
<cmath>P_{(1,2)}=\frac{1}{4}+\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}</cmath><br />
<cmath>P_{(2,2)}=\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}</cmath><br />
<cmath>P_{(1,1)}=\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}</cmath><br />
<cmath>P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}</cmath><br />
We have a system of <math>4</math> equations in <math>4</math> variables, so we can solve for each of these probabilities. Plugging the second equation into the fourth equation gives <br />
<cmath>P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}\left(\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\right)</cmath><br />
<cmath>P_{(2,1)}=\frac{8}{7}\left(\frac{1}{2}P_{(1,1)}+\frac{1}{8}P_{(1,2)}\right)=\frac{4}{7}P_{(1,1)}+\frac{1}{7}P_{(1,2)}</cmath><br />
Plugging in the third equation into this gives <br />
<cmath>P_{(2,1)}=\frac{4}{7}\left(\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\right)+\frac{1}{7}P_{(1,2)}</cmath><br />
<cmath>P_{(2,1)}=\frac{7}{6}\left(\frac{1}{7}+\frac{2}{7}P_{(1,2)}\right)=\frac{1}{6}+\frac{1}{3}P_{(1,2)}\text{ (*)}</cmath><br />
Next, plugging in the second and third equation into the first equation yields <br />
<cmath>P_{(1,2)}=\frac{1}{4}+\frac{1}{2}\left(\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\right)+\frac{1}{4}\left(\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\right)</cmath><br />
<cmath>P_{(1,2)}=\frac{3}{8}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}</cmath><br />
Now plugging in (*) into this, we get <br />
<cmath>P_{(1,2)}=\frac{3}{8}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}\left(\frac{1}{6}+\frac{1}{3}P_{(1,2)}\right)</cmath><br />
<cmath>P_{(1,2)}=\frac{3}{2}\cdot\frac{5}{12}=\boxed{\textbf{(B) }\frac{5}{8}}</cmath><br />
-mathisawesome2169<br />
<br />
==Solution 5 (Very fast)==<br />
<br />
We can immediately note that the probability of landing on any lattice point is equal to the probability of landing on another. With this in mind, we see that each vertical "boundary" has 5 lattice points. (Remember, the "corners" count!) There are two boundary lines, so there are <math>2 \times 5 = 10</math> lattice points on our desired vertical boundary lines. The total amount of lattice points on the <math>4 \times 4</math> boundary is <math>16</math>. Using <math>\frac{P(desired)}{p(total)}</math>, we get <math>\frac{10}{16} = \boxed{\textbf{(B) }\frac{5}{8}}</math> <cmath>\phantom{}</cmath><br />
-hansenhe<br />
<br />
==Video Solution 1==<br />
[https://www.youtube.com/watch?v=ZGwAasE32Y4&t=280s IceMatrix's Solution (Starts at 4:40)]<br />
<br />
==Video Solution 2==<br />
<br />
https://youtu.be/qNaN0BlIsw0<br />
<br />
==Video Solution 3==<br />
On The Spot STEM<br />
<br />
https://youtu.be/xGs7BjQbGYU<br />
<br />
==Video Solution 4==<br />
https://youtu.be/0m4lbXSUV1I<br />
<br />
~savannahsolver<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2020|ab=A|num-b=10|num-a=12}}<br />
{{AMC10 box|year=2020|ab=A|num-b=12|num-a=14}}<br />
<br />
[[Category:Introductory Probability Problems]]<br />
{{MAA Notice}}</div>Hansenhehttps://artofproblemsolving.com/wiki/index.php?title=User:OlympusHero&diff=130864User:OlympusHero2020-08-06T22:30:07Z<p>Hansenhe: /* User Count */</p>
<hr />
<div>OlympusHero's Page:<br />
<br><br />
__NOTOC__<div style="border:2px solid black; -webkit-border-radius: 10px; background:#dddddd"><br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">User Count</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black">If this is your first time visiting this page, edit it by incrementing the user count below by one.</font></div><br />
<center><font size="101px">28</font></center><br />
</div><br />
<div style="border:2px solid black; background:#cccccc;-webkit-border-radius: 10px; align:center"><br />
<br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">About Me</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black">OlympusHero is currently borderline AIME.<br><br />
<br />
OlympusHero is 10 years old.<br><br />
<br />
OlympusHero scored 38/46 when mocking the 2019 MATHCOUNTS State test, and got silver on the 2020 online MATHCOUNTS State.<br><br />
<br />
OlympusHero is a pro at maths and chess<br />
<br />
OlympusHero for black mop 2026<br />
<br />
OlympusHero was #1 at the 2017 Chess World Cadets U8 (He was 7 years old)<br />
<br />
OlympusHero is pro<br />
</font></div><br />
</div><br />
<div style="border:2px solid black; background:#bbbbbb;-webkit-border-radius: 10px; align:center"><br />
<br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Goals</div></font>==<br />
<div style="margin-left: 10px; margin-right: 10px; margin-bottom:10px">A User Count of 300<br />
<br />
Make AIME 2021 (Currently borderline)<br />
<br />
Get 5 or more on AIME I 2021 (Mocked a 3 on the 2020 AIME I a few months ago)<br />
<br />
Get to 40 or more on the MATHCOUNTS Trainer Nationals Level (25/40)<br />
</div><br />
</div></div>Hansenhehttps://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=127287User:Piphi2020-07-02T17:30:12Z<p>Hansenhe: /* User Count */</p>
<hr />
<div>{{User:Piphi/Template:Header}}<br />
<br><br />
__NOTOC__<div style="border:2px solid black; -webkit-border-radius: 10px; background:#dddddd"><br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">User Count</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black">If this is your first time visiting this page, edit it by incrementing the user count below by one.</font></div><br />
<center><font size="101px">225</font></center><br />
</div><br />
<div style="border:2px solid black; background:#cccccc;-webkit-border-radius: 10px; align:center"><br />
<br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">About Me</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black">Piphi is legendary and made the USA IMO team in 2019.<br><br />
<br />
Piphi is the creator of the [[User:Piphi/Games|AoPS Wiki Games by Piphi]], the future of games on AoPS.<br><br />
<br />
Piphi started the signature trend at around May 2020.<br><br />
<br />
Piphi is an extremely OP person - LJCoder619. <br><br />
<br />
Piphi is OP --[[User:Aray10|Aray10]] ([[User talk:Aray10|talk]]) 23:22, 17 June 2020 (EDT) <br><br />
<br />
Piphi has been very close to winning multiple [[Greed Control]] games, piphi placed 5th in game #18 and 2nd in game #19. Thanks to piphi, Greed Control games have started to be kept track of. Piphi made a spreadsheet that has all of Greed Control history [https://artofproblemsolving.com/community/c19451h2126208p15569802 here].<br><br />
<br />
Piphi also found out who won [[Reaper]] games #1 and #2 as seen [https://artofproblemsolving.com/community/c19451h1826745p15526330 here].<br><br />
<br />
Piphi has been called op by many AoPSers, including the legendary [[User:Radio2|Radio2]] himself [https://artofproblemsolving.com/community/c19451h1826745p15526800 here].<br><br />
<br />
Piphi created the [[AoPS Administrators]] page, added most of the AoPS Admins to it, and created the scrollable table.<br><br />
<br />
Piphi has also added a lot of the info that is in the [[Reaper Archives]].<br><br />
<br />
Piphi has a side-project that is making the Wiki's [[Main Page]] look better, you can check that out [[User:Piphi/AoPS Wiki|here]].<br><br />
<br />
Piphi published Greed Control Game 19 statistics [https://artofproblemsolving.com/community/c19451h2126212 here].</font></div><br />
</div><br />
<div style="border:2px solid black; background:#bbbbbb;-webkit-border-radius: 10px; align:center"><br />
<br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Goals</div></font>==<br />
<div style="margin-left: 10px; margin-right: 10px; margin-bottom:10px">A User Count of 330<br />
{{User:Piphi/Template:Progress_Bar|66.67|width=100%}}<br />
<br />
200 subpages of [[User:Piphi]]<br />
{{User:Piphi/Template:Progress_Bar|38.5|width=100%}}<br />
<br />
200 signups for [[User:Piphi/Games|AoPS Wiki Games by Piphi]]<br />
{{User:Piphi/Template:Progress_Bar|100|width=100%}}<br />
<br />
Make 10,000 edits<br />
{{User:Piphi/Template:Progress_Bar|13.23|width=100%}}</div><br />
</div></div>Hansenhehttps://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=127286User:Piphi2020-07-02T17:30:01Z<p>Hansenhe: /* User Count */</p>
<hr />
<div>{{User:Piphi/Template:Header}}<br />
<br><br />
__NOTOC__<div style="border:2px solid black; -webkit-border-radius: 10px; background:#dddddd"><br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">User Count</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black">If this is your first time visiting this page, edit it by incrementing the user count below by one.</font></div><br />
<center><font size="101px">230</font></center><br />
</div><br />
<div style="border:2px solid black; background:#cccccc;-webkit-border-radius: 10px; align:center"><br />
<br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">About Me</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black">Piphi is legendary and made the USA IMO team in 2019.<br><br />
<br />
Piphi is the creator of the [[User:Piphi/Games|AoPS Wiki Games by Piphi]], the future of games on AoPS.<br><br />
<br />
Piphi started the signature trend at around May 2020.<br><br />
<br />
Piphi is an extremely OP person - LJCoder619. <br><br />
<br />
Piphi is OP --[[User:Aray10|Aray10]] ([[User talk:Aray10|talk]]) 23:22, 17 June 2020 (EDT) <br><br />
<br />
Piphi has been very close to winning multiple [[Greed Control]] games, piphi placed 5th in game #18 and 2nd in game #19. Thanks to piphi, Greed Control games have started to be kept track of. Piphi made a spreadsheet that has all of Greed Control history [https://artofproblemsolving.com/community/c19451h2126208p15569802 here].<br><br />
<br />
Piphi also found out who won [[Reaper]] games #1 and #2 as seen [https://artofproblemsolving.com/community/c19451h1826745p15526330 here].<br><br />
<br />
Piphi has been called op by many AoPSers, including the legendary [[User:Radio2|Radio2]] himself [https://artofproblemsolving.com/community/c19451h1826745p15526800 here].<br><br />
<br />
Piphi created the [[AoPS Administrators]] page, added most of the AoPS Admins to it, and created the scrollable table.<br><br />
<br />
Piphi has also added a lot of the info that is in the [[Reaper Archives]].<br><br />
<br />
Piphi has a side-project that is making the Wiki's [[Main Page]] look better, you can check that out [[User:Piphi/AoPS Wiki|here]].<br><br />
<br />
Piphi published Greed Control Game 19 statistics [https://artofproblemsolving.com/community/c19451h2126212 here].</font></div><br />
</div><br />
<div style="border:2px solid black; background:#bbbbbb;-webkit-border-radius: 10px; align:center"><br />
<br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Goals</div></font>==<br />
<div style="margin-left: 10px; margin-right: 10px; margin-bottom:10px">A User Count of 330<br />
{{User:Piphi/Template:Progress_Bar|66.67|width=100%}}<br />
<br />
200 subpages of [[User:Piphi]]<br />
{{User:Piphi/Template:Progress_Bar|38.5|width=100%}}<br />
<br />
200 signups for [[User:Piphi/Games|AoPS Wiki Games by Piphi]]<br />
{{User:Piphi/Template:Progress_Bar|100|width=100%}}<br />
<br />
Make 10,000 edits<br />
{{User:Piphi/Template:Progress_Bar|13.23|width=100%}}</div><br />
</div></div>Hansenhehttps://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=126283User:Piphi2020-06-23T18:16:15Z<p>Hansenhe: /* User Count */</p>
<hr />
<div>{{User:Piphi/Template:Header}}<br />
<br><br />
__NOTOC__<div style="border:2px solid black; -webkit-border-radius: 10px; background:#dddddd"><br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">User Count</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black">If this is your first time visiting this page, edit it by incrementing the user count below by one.</font></div><br />
<center><font size="101px">200</font></center><br />
</div><br />
<div style="border:2px solid black; background:#cccccc;-webkit-border-radius: 10px; align:center"><br />
<br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">About Me</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black">Piphi is legendary and made the USA IMO team in 2019.<br><br />
<br />
Piphi is the creator of the [[User:Piphi/Games|AoPS Wiki Games by Piphi]], the future of games on AoPS.<br><br />
<br />
Piphi started the signature trend at around May 2020.<br><br />
<br />
Piphi is an extremely OP person - LJCoder619. <br><br />
<br />
Piphi is OP --[[User:Aray10|Aray10]] ([[User talk:Aray10|talk]]) 23:22, 17 June 2020 (EDT) <br><br />
<br />
Piphi has been very close to winning multiple [[Greed Control]] games, piphi placed 5th in game #18 and 2nd in game #19. Thanks to piphi, Greed Control games have started to be kept track of. Piphi made a spreadsheet that has all of Greed Control history [https://artofproblemsolving.com/community/c19451h2126208p15569802 here].<br><br />
<br />
Piphi also found out who won [[Reaper]] games #1 and #2 as seen [https://artofproblemsolving.com/community/c19451h1826745p15526330 here].<br><br />
<br />
Piphi has been called op by many AoPSers, including the legendary [[User:Radio2|Radio2]] himself [https://artofproblemsolving.com/community/c19451h1826745p15526800 here] (note: Radio2 calls lots of users OP).<br><br />
<br />
Piphi created the [[AoPS Administrators]] page, added most of the AoPS Admins to it, and created the scrollable table.<br><br />
<br />
Piphi has also added a lot of the info that is in the [[Reaper Archives]].<br><br />
<br />
Piphi has a side-project that is making the Wiki's [[Main Page]] look better, you can check that out [[User:Piphi/AoPS Wiki|here]].<br><br />
<br />
Piphi published Greed Control Game 19 statistics [https://artofproblemsolving.com/community/c19451h2126212 here].</font></div><br />
</div><br />
<div style="border:2px solid black; background:#bbbbbb;-webkit-border-radius: 10px; align:center"><br />
<br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">[[User:Piphi/Asymptote|Asymptote]]</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px">For a complete list of my Asymptote drawings, go [[User:Piphi/Asymptote|here]].</div><br />
</div></div>Hansenhe